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Natalka [10]
3 years ago
10

If the pressure of a gas is 2.000 × 104 kPa, what is its pressure in atm?

Chemistry
2 answers:
lisov135 [29]3 years ago
7 0

Answer:

197200atm

Explanation:

There are different units to express pressure of a gas, some of them are atmosphere, Bar, Kilopascal, Pascal, torr among other.

In this exercise we need to know the equivalence between atm and Kpa.

                                             if, 1kp=0,00986atm

then,

(2,000 x104kPa)*(0,00986atm)/1Kpa= 0,00986 atm

luda_lava [24]3 years ago
4 0

<u>Answer:</u>

<em>197.4 atm is the Answer.</em>

<em></em>

<u>Explanation:</u>

Pressure can be expressed in unit of torr, mm Hg, psi, atm, kPa etc

<em>1 atm = 760 torr = 760 mm Hg = 14.7 psi = 101.325 kPa </em>

So Here to convert atm to kPa the conversion factor is either \frac {1atm}{101.325kPa} or \frac {101.325kPa}{1atm}

We need the answer in atm so kPa should get cancel and it should be in the denominator

2.000 \times 10^4 kPa\times \frac {1atm}{101.325kPa}

=197.4 atm is the Answer.

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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
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There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

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<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0
3 years ago
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